What are the solutions to the equation 32y^2+32y+8=0 ?

The solutions to the equation 32y^2+32y+8=0 are y=-1/2

Find the zeros of 32y^2+32y+8=0

The zeros of 32y^2+32y+8=0 are y=-1/2

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32 y^{2} + 32 y + 8 = 0

y = - \frac{1}{2}

Decimal Notation

y = - 0.5

Solution steps

32 y^{2} + 32 y + 8 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 32 \pm 3 2 ^{2} - 4 \cdot 32 \cdot 8}{2 \cdot 32}

3 2^{2} - 4 \cdot 32 \cdot 8 = 0

y_{1, 2} = \frac{- 32 \pm 0}{2 \cdot 32}

y = \frac{- 32}{2 \cdot 32}

\frac{- 32}{2 \cdot 32} = - \frac{1}{2}

y = - \frac{1}{2}

The solution to the quadratic equation is:

y = - \frac{1}{2}

x^{2} = \frac{lo g _{10} ( 25 )}{lo g _{10} ( 2 )}

x, x^{2} - 2 x y - y^{2} + x - 12 y = 0

x, 16 x^{2} - y^{2} + 16 z^{2} = 16

x, x^{2} - 2 x + 4 = 0

(x - 2)^{2} = 20

What are the solutions to the equation 32y^2+32y+8=0 ?
The solutions to the equation 32y^2+32y+8=0 are y=-1/2
Find the zeros of 32y^2+32y+8=0
The zeros of 32y^2+32y+8=0 are y=-1/2